Recently I studied the orbit stabilizer theorem which is as follows:

Suppose $G$ is a group acting on $X$ (i.e.$X$ is a $G$-set). Let $x\in X$,then define, $\operatorname {orb}(x):=\{g.x:g\in G\}$ and $\operatorname{stab}(x)=\{g\in G:g.x=x\}$, then we have $|\operatorname{orb}(x)|=[G:\operatorname{stab}(x)]$, provided $G$ and $X$ are finite.

Now I have tried to intuitively understand this theorem like this, orbit of $x$ is roughly speaking, all the possible points in $X$ where $x$ can go under the given group action. And stabilizer means all the group elements(we can think of them as permutations also) that fix $x$.

Now the index on the right hand side of orbit stabilizer theorem is the number of cosets of $G$ induced by the subgroup viz stabilizer of $x$. Whenever a subgroup induces a coset or partition within a group, it means that we are classifying the group element by discriminating the group elements based on the propertly of that subgroup, now the property of stabilizer of $x$ is that it fixes $x$, so it contains all elements of $G$ that fix $x$, so the other coset must be according to the property,where the group elements take $x$, all the members of the group taking $x$ to a specific point will be a coset and for every point in orbit, there is a coset of $G$ whose members take $x$ to that point.

So there is a one-one correspondence between the cosets of $G$ induced by stabilizer subgroup and the members of the orbit of $x$,notice that $x$ itself is a member of the orbit and for it,the corresponding coset is stabilizer of $x$ itself. So,naturally the number of orbit elements is the same as the number of cosets of stabilizer of $x$ in $G$.

I want to know if my intuition or understanding correct?

  • $\begingroup$ @Berci can you give me suitable diagram. $\endgroup$ Feb 2, 2020 at 13:42
  • $\begingroup$ Not a diagram, but you can draw one based on my answer. $\endgroup$
    – Berci
    Feb 2, 2020 at 13:48
  • 1
    $\begingroup$ I've always been a big fan of this blog post by Timothy Gowers. It starts off with a slightly more concrete intuitive example of orbit-stabiliser, and goes on to formalise this intuition. Perhaps you will find it helpful. $\endgroup$ Feb 2, 2020 at 13:59
  • $\begingroup$ @IzaakvanDongen I wanted to know one more thing $Ker\phi=\cap_{x\in X} stab(x)$ right?where $ker\phi$ is the kernel of $\phi : G \to Sym(X)$ defined by $\phi(g)=\sigma_g$ where $\sigma_g(x)=(g,x)$ where $(.,.)$ is a group action of $G$ on $X$. $\endgroup$ Feb 2, 2020 at 14:02
  • $\begingroup$ Yes, in order to be in $\operatorname{Ker} \phi$ you should fix everything, and hence be in every stabiliser, and conversely if you are in every stabiliser then you fix everything so your permutation representation is the identity permutation. $\endgroup$ Feb 2, 2020 at 14:04

1 Answer 1


Yes, it is correct.

The (left) cosets of $S:=\mathrm{stab}(x)$ correspond bijectively to the elements in the orbit of $x$:

Specifically, for any elements $g,h\in G$, we have $$gx=hx\ \iff\ h^{-1}gx=x\ \iff\ h^{-1}g\in S\ \iff\ gS=hS\,.$$


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